Problem: Solve for $z$, $ \dfrac{3}{4z^2} = -\dfrac{4}{2z^2} - \dfrac{4z - 6}{2z^2} $
First we need to find a common denominator for all the expressions. This means finding the least common multiple of $4z^2$ $2z^2$ and $2z^2$ The common denominator is $4z^2$ The denominator of the first term is already $4z^2$ , so we don't need to change it. To get $4z^2$ in the denominator of the second term, multiply it by $\frac{2}{2}$ $ -\dfrac{4}{2z^2} \times \dfrac{2}{2} = -\dfrac{8}{4z^2} $ To get $4z^2$ in the denominator of the third term, multiply it by $\frac{2}{2}$ $ -\dfrac{4z - 6}{2z^2} \times \dfrac{2}{2} = -\dfrac{8z - 12}{4z^2} $ This give us: $ \dfrac{3}{4z^2} = -\dfrac{8}{4z^2} - \dfrac{8z - 12}{4z^2} $ If we multiply both sides of the equation by $4z^2$ , we get: $ 3 = -8 - 8z + 12$ $ 3 = -8z + 4$ $ -1 = -8z $ $ z = \dfrac{1}{8}$